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If you pick any $3$ numbers in the given order from the array(sorted) remaining $5$ elements are already sorted. You can not change the relative position of those $5$ elements because they are distinct and already sorted. So no of ways $= {}^8C_3.$

Same argument holds for picking up $5$ elements initially. No of ways $= {}^8C_5.$

here the question asks only for selection and not for an arrangement of the selected nos. This is where I got confused. pick 3 or 5 elements, remaining elements will remain sorted since all are distinct nos.

The question actually is based on combinations not merging.Firstl y two arrays are taken : B having 3 elements and C having 5 elements sorted in ascending order and then merged into a single array A having 8 elements.We have to find no of distinct sequences. 8 elements can be arranged in 8! ways out of that 3 elements of array B can be arranged in 3! ways and 5 elements of C can be arranged in 5! ways. Now, No. of distinct sequences = 8!/3! 5! = 56 Ans.